Problem Solving Strategies

Wolgwang

6:09 AM

@JohnRennie Hi!

@Wolgwang hi :-)

Are you free?

Yes

> A person is standing on a road by holding an umbrella at right angle to the horizontal in order to protect himself from rain falling at speed of 10m/s,and where as the velocity of wind is 0. Suddenly wind blows at a speed of 20m/s, towards 30degree south of west. Now by what angle the person has to turn the umbrella to protect himself from rain?

$$V_{Init.rain}=-10\hat k\\V_w=-10\sqrt3 \hat{i} -10\hat j\\V_{Finalrain}=-10\sqrt3 \hat{i} -10\hat j-10\hat k$$

What to do next?

John Rennie

You just need to find the angle of the vector along which the rain moves. Yes?

Wolgwang

6:13 AM

Yes

John Rennie

I don't think towards 30degree south of west makes any difference.

Wolgwang

Hmm?

John Rennie

The important point is that the wind is blowing horizontally at 20 m/s and the rain is falling vertically at 10 m/s

So the angle the rain makes with the horizontal is tan θ = 10/20

I can draw a diagram if it will help ...

Wolgwang

Yes :-)

John Rennie

@Wolgwang the rain falls down through the air at 10 m/s, and the air is moving horizontally at 20 m/s. Yes?

Wolgwang

6:22 AM

@JohnRennie Yes

John Rennie

So the rain travels along the red vector.

Wolgwang

Yes

John Rennie

And you need to turn your umbrella so it points along the red vector.

Wolgwang

Yes

John Rennie

Since you started with the umbrella pointing straight up you need to turn your umbrella by an angle of 90 - θ. Yes?

Wolgwang

6:24 AM

But I also need to turn the umbrella on the z-axis.

John Rennie

Umbrellas are axially symmetric. An umbrella has a top and bottom, but not a front or back. So you do not need to rotate the umbrella about the z axis.

That's why towards 30degree south of west makes no difference.

Wolgwang

:-o Didn't know that

John Rennie

Have you never seen an umbrella?

Wolgwang

I have

John Rennie

Well if you are holding an umbrella straight up, i.e. along the z axis, then rotating the umbrella about this axis doesn't change it. Yes?

Wolgwang

6:29 AM

What does rotating about the axis mean?

Changing vector's head in xy?

John Rennie

Here is a top down view of an umbrella I found on Google images:

So we are looking straight downwards and you are sheltering underneath the umbrella.

i.e. we are looking down the z axis. Yes?

Ok I got it

:-)

Thanks :-)

@JohnRennie Hii sir

I have quite a few questions ask.

Are you free rn?

Whenever you are. This is the Q

Q1 Now , average speed of particles = $\frac{\sqrt{8RT}}{\sqrt{ \pi*M}}$ where M is Molar mass of the gas.

Pressure by real gases : $\frac{RnT}{V-(nb)}$ - $\frac{n^2}{v^2}$-a . Here , a is strength of attraction between particles and b is volume occupied by mass of 1 mole particle.

So , My Q is that why don’t we consider in either of the formula the repulsion forces,collision between the particles ? Like , it is not mentioned anywhere in these equations.

Q2 is this Here. There are two gases He both.

Volume of above is 8L and below is 2L

Left side diagram is Q

We have founded out the pressure for both the gases on right

Then subtracted them

Now , what I got earlier was that the below piston will surely move up because we also say that the temperature is increasing gradually

It keeps on increasing

Coming back , P of gas below piston - P of gas above piston = weight of piston / Area of piston? This is what I didn’t get. Why the weight of piston ?

There are two Q here. We can with anyone first

I am here. Just ping me when we can discuss sir

John Rennie

6:43 AM

@SrijanM.T hi, yes I'm free.

The equation $v_{rms} = \frac{\sqrt{8RT}}{\sqrt{ \pi*M}}$ is for an ideal gas.

Srijan M.T

Yes sir

John Rennie

So in that case there are no interactions between the gas molecules.

If there are interactions then the equation becomes a lot more complicated.

Srijan M.T

Ohk.Accepted for ideal gas

@JohnRennie Right

What about real gas scenario

the equation I wrote second one

John Rennie

For an ideal gas $PV = nRT$. Yes?

Srijan M.T

Yes

John Rennie

6:46 AM

For a real gas the equation will be much more complicated, but we can make some approximations.

Srijan M.T

Ohk.

I wrote it sir. For real gas the equation

John Rennie

The volume is the volume that the gas molecules are free to move in, but if the gas molecules are not points (like in an ideal gas) then they take up some volume $b$.

Srijan M.T

@JohnRennie Yes sir

John Rennie

So the volume the gas molecules can move in is $V-b$. So we can change our equation to:

$$ P(V-b) = nRT$

Srijan M.T

Yes

John Rennie

6:48 AM

For some constant $b$ that depends on the size of the gas molecules.

Srijan M.T

Yes.

John Rennie

And now suppose there is an attractive force between the gas molecules.

Srijan M.T

Ohk.

Since they have mass. So , there is attraction

John Rennie

Since this pulls the gas molecules together it will reduce the pressure the gas molecules exert on the container they are in. Yes?

Srijan M.T

Yes

John Rennie

6:49 AM

So the pressure is going to be reduced a bit to $P-x$ where $x$ is some value.

Srijan M.T

Yes

John Rennie

It turns out that a good choice for $x$ is $a/V_m^2$ where $V_m$ is the molar volume.

Srijan M.T

Ohk.

John Rennie

And $V_m = V/n$ i.e. volume per mole, so that makes $x = an^2/V^2$

And our equation becomes:

Srijan M.T

Yes sir. Right

John Rennie

6:52 AM

$$ \left( P - \frac{an^2}{V^2} \right) (V - b ) = nRT $$

Srijan M.T

Yes sir right

John Rennie

But this is an approximation.

Srijan M.T

Ohk.Yes

John Rennie

Real gases would be far more complicated.

Srijan M.T

@JohnRennie K

John Rennie

6:53 AM

That approximation gives a reasonably good description for a real gas like nitrogen or oxygen.

Srijan M.T

We considered attraction forces I,e the value “a”

@JohnRennie K

John Rennie

And it's called the Van der Waal's equation of state.

Srijan M.T

But what about repulsion , collision forces for it. Since real gas particles have mass

@JohnRennie Yes sir, ok

John Rennie

@SrijanM.T yes, $a$ is a constant that describes the attraction. The repulsion is included in the constant $b$.

Srijan M.T

Sir , b is the amount of volume occupied by 1 mole particles right ?

John Rennie

6:55 AM

No, $b$ is approximately the volume occupied by a mole of gas molecules, but only approximately.

Srijan M.T

Ohk.

But sir , how does this b tell us about repulsion force ?

John Rennie

In practice we have to make measurements on a real gas and calculate the values of a and b from those measurements.

Srijan M.T

It talks about volume right

John Rennie

Repulsion means it's hard to squeeze two molecules together. Yes?

Srijan M.T

I am thinking repulsion as when two molecules collide with each other and then go away from each with more velocity

John Rennie

6:57 AM

If it wasn't hard to squeeze two molecules together they would never collide because they could get arbitrarily close to each other.

So collisions occur because of repulsive forces. Yes?

Srijan M.T

Yes

John Rennie

So:

2 mins ago, by John Rennie

Repulsion means it's hard to squeeze two molecules together. Yes?

Srijan M.T

Ohk. Got it

@JohnRennie I can do the maths after that.

Got this Q. Shall we move to 2nd one now ?

John Rennie

OK :-)

Q2 is this Here. There are two gases He both.
Volume of above is 8L and below is 2L
Left side diagram is Q
We have founded out the pressure for both the gases on right
Then subtracted them
Now , what I got earlier was that the below piston will surely move up because we also say that the temperature is increasing gradually
It keeps on increasing
Coming back , P of gas below piston - P of gas above piston = weight of piston / Area of piston? This is what I didn’t get. Why the weight of piston ?
https://i.stack.imgur.com/rQXwf.jpg

Yes sir

7:02 AM

Yes sir.

Give me a moment to read through the question.

Srijan M.T

Ohk. Sure

John Rennie

> Coming back , P of gas below piston - P of gas above piston = weight of piston / Area of piston? This is what I didn’t get. Why the weight of piston ?

Call the pressure above P₁ and the pressure below P₂

So if the area of the piston is A then the gas above it exerts a downwards force on the piston P₁A and the gas below exerts an upwards force P₂A. The net upwrds force on the piston due to the gas is Fup = (P₂ - P₁)A.

OK so far?

Srijan M.T

One sec

Yes sir right

John Rennie

7:08 AM

And the weight of the piston exerts a downwards force Fdown = mg. Yes?

Srijan M.T

Yes

John Rennie

So the net force on the piston is:

Fnet = Fup - Fdown = (P₂ - P₁)A - mg

Srijan M.T

OH yes

I got it

@JohnRennie Amazing

John Rennie

:-)

Srijan M.T

Sir , if you’re free. I have one more Q

John Rennie

7:09 AM

OK ... ?

Srijan M.T

More of problem solving

1 min. I’ll paste it

John Rennie

OK, how far did you get with this?

Srijan M.T

Two chambers having certain PVT. After stopper is removed , Q is to find Final pressure on both sides

@JohnRennie Very far. I will explain you the solution . Pls correct me

John Rennie

OK ...

Srijan M.T

There are two cases here. One is when T is constant even though pressure changes and other one is when T is not constant

Solution for case 1:

When the stopper is removed , our final conclusion is that pressure is same on both sides. Ok ?

John Rennie

7:14 AM

Yes

Srijan M.T

So , let us assume no of moles on left side = L and on RHS = R

John Rennie

Yes, that's how I would start.

Srijan M.T

Now , assuming that no of mole particles travel more RHS to LHS. Since particles flow from high pressure to low pressure. It is just an assumption that RHS has more particles but in end of our ans comes -ve , then we know that transition happening from LHS to RHS

Ok so far ?

John Rennie

Yes

Srijan M.T

@JohnRennie Feel good to know that

Now , we can say

R - x( x means the no of particles that went from RHS to LHS)

John Rennie

7:17 AM

I think you may be making this more complicated than it needs to be ...

Srijan M.T

and L+ x( means that those x particles have traveled to LHS)

@JohnRennie There are two methods I will solve it with

This one is lengthy. Not second one

John Rennie

OK

Srijan M.T

Just bear with me till 2nd one. After that , I’ll discuss with you

John Rennie

OK

Srijan M.T

R(L+x) T1/V1 = R(n_R+x)(T2)/V2

The R in bracket for 2nd one means moles

So , by doing this. We can calculate x and then in end , find P final

So , this is 1 method

Pls check if it is correct

John Rennie

7:20 AM

OK. Though you cannot assume the temperatures stay constant ...

Srijan M.T

@JohnRennie Yes. But here , it is given in Q to assume them to be cosnt

John Rennie

OK

Srijan M.T

In case 2 when we do , I won’t assume that

Ok. Now , METHOD 2, not. CASE 2

Now , let us say n_L = 20 and n_R = 10

After solving , just keeping simple values for now

Ok so far ?

John Rennie

Yes

Srijan M.T

Now , I Cassie that molecules move from RHS to LHS. So , since the temp is constant. The molecules from RHS to LHS cool down by 300K

Due to temp difference

We can say it cools down by 2/3rd factor

Ok so far ?

John Rennie

7:25 AM

OK

By ²⁄₃ or by ¹⁄₃?

Srijan M.T

2/3

It becomes 600K on right

900*2/3

Since P= nRT/V

So T changes by 2/3rd

Ok so far ?

John Rennie

OK

Srijan M.T

So , I can 10-X from RHS

But for LHS , Volume has decreased when they entered into LHS

Also , P=nRT/V

So , if T changes by 2/3rd factor , then also P changes by same factor

So , it will be 20+X(2/3)

Ok so far ?

It took me hours to think about this method . I hope it’s ok what I have done till now

There is little more

John Rennie

OK ...

Srijan M.T

So , now . We know P is directly proportional to V

That has change of 3rd factor

So , in final it becomes 20+X(2/3)(3)=10-X

I knew about this method till constant volume and temp. I thought to take not constant volume , so I thought of this then.

Ok so far? Done now sir

I hope this is the one you were thinking too

John Rennie

7:35 AM

I as thinking of the situation where the temperature equilibrates so it's equal on both sides, but if the question says to keep the temperatures unchanged that wouldn't apply.

Srijan M.T

@JohnRennie ohk.

Q says that even after changing moles , T is 600 and 900 K only

On LHS and rhs

John Rennie

I guess what I'd say is to use P = nRT/V

Srijan M.T

Hmm K. The only problem with that is that it can become lengthy

John Rennie

So start by calculating n₁ and n₂ before the tap is opened.

Srijan M.T

@JohnRennie Yes. That is right. In Q where values like R is there and then V is also difficult value. It can be lengthy to find nL

and nR

John Rennie

7:38 AM

And after the tap is opened the amount of gas will change to N₁ and N₂, but the total amount of gas has to stay the same so N₁ and N₂ = n₁ + n₂

And we know with the tap open P₁ = P₂

Srijan M.T

What is N?

Is it V?

John Rennie

Number of moles after the tap is opened.

Srijan M.T

N1 and n1 are same ?

John Rennie

I called it N to distinguish it from the number of moles before the tap is opened n

Srijan M.T

Ohk.

@JohnRennie Yes

Sir. Shall we assume T is not constant for this case. That is what I am finding difficult to solve

T1=T2 here. Unlike the 1st one

John Rennie

7:41 AM

So after the tap is open N₁RT₁/V₁ = N₂RT₂/V₂. Yes?

Srijan M.T

Yes. Any difference if T1=T2 or T1 is not equal to T2 ?

In the equation

Oh. I think T will get cut

John Rennie

This is always true, but if we allow the temperatures to change we no longer know the values of T₁ and T₂.

Just to be clear, what calculation are we trying to do? Constant temp or changing temp?

Srijan M.T

Yes. So , it should be T on both sides right. Just like we considered P final of both to be same

John Rennie

And if it's changing temp are we assuming the temp is the same on both sides?

Srijan M.T

@JohnRennie I want to do changing sir. I was confused about it a bit

@JohnRennie Yes

@JohnRennie I think so right ?

John Rennie

7:45 AM

If we allow the temp to equilibrate to be the same on both sides then it's easy because then we have P (V₁ + V₂) = (n₁ + n₂)RT. Yes?

Srijan M.T

Sir. Are we assuming both the chambers as one system ?

John Rennie

Yes. With the tap open it's one system because the gas molecules can move freely between each side.

Srijan M.T

Ohk. So , I was correct about T assumption

Like it will get cut on both sides

N₁RT/V₁ = N₂RT/V₂.

John Rennie

And we know the internal energy has not changed because no heat was added and no work was done.

Srijan M.T

@JohnRennie ohk. Right

Sir, I have one confusion. How is N1 AND N2 both n1+n2

John Rennie

7:49 AM

So that means n₁CT₁ + n₂CT₂ = (n₁ + n₂)CT.

Srijan M.T

One has to be n1-x and other n2+x right ?

John Rennie

@SrijanM.T forget about the uppercase N's. That was for a different calcualion.

46 secs ago, by John Rennie

So that means n₁CT₁ + n₂CT₂ = (n₁ + n₂)CT.

Srijan M.T

Ok. C is heat capacity ?

John Rennie

Yes

Srijan M.T

Ohk.

@JohnRennie So state 1 + state 2 = final state ?

John Rennie

7:50 AM

So T = (n₁T₁ + n₂T₂)/(n₁ + n₂)

Srijan M.T

@JohnRennie Didn’t get this one sir ?

John Rennie

@SrijanM.T I'm just saying the internal energy before the tap is opened is Uinit = n₁CT₁ + n₂CT₂ and after the tap is opened and the two sides have equilibrated Ufinal = (n₁ + n₂)CT where T is the final temperature. Yes?

Srijan M.T

Yes

John Rennie

And the internal energy is constant so Uinit = Ufinal. Yes?

Srijan M.T

I don’t know about internal energy actually. Sorry. I am yet to do thermo. My only Q Uinital = state 1 + state 2 means that both the states are considered as one system

So , final system is also kind of line one system

Because in the end, also the final system is same on both sides .

@JohnRennie So , I got it.

Sir @JohnRennie ? Are you there

John Rennie

7:58 AM

So Uinit = Ufinal and substituting for Uinit = n₁CT₁ + n₂CT₂ and Ufinal = (n₁ + n₂)CT we get:

n₁CT₁ + n₂CT₂ = (n₁ + n₂)CT

Yes?

Srijan M.T

Yes

John Rennie

And that rearranges to:

T = (n₁T₁ + n₂T₂)/(n₁ + n₂)

And we know everything on the right hand side so we can calculate the final temperature T.

OK so far?

Srijan M.T

Ohk

John Rennie

Then we just use:

P (V₁ + V₂) = (n₁ + n₂)RT

To calculate the final pressure P.

Srijan M.T

Yes

John Rennie

8:00 AM

And that's it.

Srijan M.T

Pretty interesting.

Amazing one sir.

I got it too.

John Rennie

:-)

Srijan M.T

ohk sir. Thanks a lot. See you soon.

John Rennie

Bye :-)

Srijan M.T

Also , I hope you’re safe and healthy. Take care sir.

Srijan M.T

9:13 AM

@JohnRennie Hii sir

1

Q: Doubt regarding proof of $pV = nRT$

In my textbook, the proof for this formula is as follows: $$F \propto \text{Area}_{\text{wall}} \cdot \text{(number of density of particles I.e here m is const) I.e} \frac{n}{V} *\text{Average Kinetic energy of particles}$$ Then, we get $$\frac{F}{A} = R(\text{Prop const}) \cdot \frac{n}{v} \c...

Do have a look at this Q

John Rennie

It isn't clear to me where the problem is.

Which of the two equations you give do you mean by equation 1?

Srijan M.T

9:29 AM

@JohnRennie Ohk

The top one which is in yellow colour background

>$$F \propto \text{Area}_{\text{wall}} \cdot \text{(number of density of particles I.e here m is const) I.e} \frac{n}{V} *\text{Average Kinetic energy of particles}$$

This

If F is directly proportional, then it means ma is directly proportional

Ok so far ?

John Rennie

So you're asking how that equation is derived?

Srijan M.T

There two things I ask kind of

One is derivation

other one is that

Can you prove it that after simplifying the equation or simplifying it dimensionally

It is equal to ma or F

Is it ok? I’ll draw and explain if not clear

John Rennie

The equation is correct, but it isn't derived directly. Or at least that isn't how I would derive it.

Srijan M.T

Yes. I want to know how to derive it.

John Rennie

Start with a cube of side L containing particles with mass m moving at velocity v.

Srijan M.T

9:35 AM

Ok

John Rennie

As a first approximation we'll assume the particles are moving parallel to one of the sides, so they are striking one wall of the cube at normal incidence.

OK so far, or should I draw a diagram?

Srijan M.T

I understood it. Just like 10 person running parallel to each other and in same direction

John Rennie

OK, now consider one molecule hitting the face of the cube. It hits the face at velocity v then bounces back with velocity -v so the change in momentum is Δp = 2mv. Yes?

Srijan M.T

Yes

John Rennie

Now, we are going to calculate the total momentum change in one second because this is dp/dt and we know F = dp/dt. So if we calculate the momentum change in one second we know the average force the molecule exerts on the face of the cube.

OK so far?

Srijan M.T

9:40 AM

one sec

Yes

Can you explain a little what this average force would mean ? Like force applied in the time interval 0-1 sec

John Rennie

OK. When the molecule hits the wall it bounces back. Then it has to travel a distance L across the cube, bounce off the opposite face then travel a distance back to the face again. So in between collisions with the face it has to travel a distance 2L. Yes?

Srijan M.T

Ohk

John Rennie

And the time it takes to travel a distance 2L is t = 2L/v, so the time in between collisions is t = 2L/v. Yes?

Srijan M.T

The blue particle I drew

John Rennie

Yes

Srijan M.T

9:44 AM

@JohnRennie Ohk

John Rennie

So the number of collisions with the face per second is 1/t and that is v/2L.

And each collision changes the momentum by 2mv

Srijan M.T

Does it travel L distance in 1sec

John Rennie

So the total change of momentum per second is v/2L x 2mv = mv²/L

Srijan M.T

or 2L in 1sec

?

John Rennie

@SrijanM.T suppose we are calculating the force on the right face. Then after the molecule collides with the right face it bounces back and starts travelling left.

Srijan M.T

9:47 AM

Yes

John Rennie

It travels a distance L then it hits the left face. When it hits the left face it bounces again and starts travelling right.

Srijan M.T

Yes

John Rennie

Then it travels another distance L before it hits the right face again.

So the distance it has to travel between collisions with the right face is 2L.

OK so far?

Srijan M.T

Yes

John Rennie

And if it is travelling at a speed v then the time it takes to travel a distance 2L is t = 2L/v

Yes?

Srijan M.T

9:49 AM

Yes

@JohnRennie I didn’t understand 1/t?

John Rennie

Suppose you hit something every 0.1 seconds, how many times a second do you hit it?

Srijan M.T

1/10 times

Right ?

John Rennie

Really? If you hit something every 0.1 seconds then the number of times you hit it in one second is 1/10?

Srijan M.T

In total 1sec = 10times

John Rennie

Yes. Because 10 = 1/0.1

Yes?

Srijan M.T

9:52 AM

Yes. Sorry , didn’t get it quick.

Understood now

John Rennie

So if the time in between collisions with the surface is t then the number of hits per second is 1/t.

And 1/t = v/2L.

Srijan M.T

Yes sir

John Rennie

And at each collision the momentum change is 2mv, so the total momentum change per second is v/2L x 2mv = mv²/L

Srijan M.T

Yes sir

John Rennie

OK. That was for one molecule. But now suppose we have n molecules in the box, then to get the total momentum change for all n molecules we just multiply by n. So we get the total momentum change in 1 second:

Δp = n mv²/L

OK so far?

Srijan M.T

9:56 AM

Yes sir. Right

John Rennie

But the momentum change per second is just the force, because F = dp/dt. Yes?

Srijan M.T

Yes

John Rennie

And pressure = F/A

Srijan M.T

Yes

Right

John Rennie

For our cube the area of the face is L² so we can calculate the pressure by dividing both sides by L². That gives:

P = n mv²/L³

Srijan M.T

9:58 AM

Ok

John Rennie

And n/L³ is the number of particles per unit volume i.e. the number density. Let's call this N. And mv² is twice the KE, so we can write our equation as:

P = N x 2KE

Srijan M.T

Hmm K. Now we’re getting it.Nice.understood

John Rennie

In your question you have multiplied both sides by the area to get the force F =PA, so you ended up with the equation:

F = A x N x 2KE

Srijan M.T

Ohk.

I feel we have everything now. Just 2KE should be KE

John Rennie

So that's how you get that equation. Your equation has a proportional symbol not equality because actually we used a really crude approximation. In a real gas the molecules are travelling in all directions at random.

Srijan M.T

10:03 AM

Ohk.

Also , They took K.E = Temp

John Rennie

So our calculation will have some correction factor that is probably complicated to calculate. We won't worry what the correction fact is. Instead we'll just say the force is proportional to A x N x KE.

@SrijanM.T that's because the average KE of an ideal gas molecule is equal to ³⁄₂kT.

Do you know why that is?

Srijan M.T

No sir. Haven’t reached that far. We did thermo a bit and then gas laws.

I don’t know the reason but only knew that first law equation for ideal gas

John Rennie

Equipartition theorem

In classical statistical mechanics, the equipartition theorem relates the temperature of a system to its average energies. The equipartition theorem is also known as the law of equipartition, equipartition of energy, or simply equipartition. The original idea of equipartition was that, in thermal equilibrium, energy is shared equally among all of its various forms; for example, the average kinetic energy per degree of freedom in translational motion of a molecule should equal that in rotational motion. The equipartition theorem makes quantitative predictions. Like the virial theorem, it gives the...

Srijan M.T

Ohk. I will check it for sure then,

John Rennie

The equipartition theorem states that at a temperature T every degree of freedom of a system gets an energy of ¹⁄₂kT.

Srijan M.T

10:06 AM

Ohk. So , there I guess the 2 of K.E gets cancelled out

John Rennie

And a single particle has three degrees of freedom corresponding to motion in the x, y and z directions.

Srijan M.T

Yes.

John Rennie

So the energy of a single particle is 3 x ¹⁄₂kT = ³⁄₂kT

And the only energy a single particle has is its KE. Hence the average KE = ³⁄₂kT

Srijan M.T

Ohk.

Satwik

@JohnRennie, my teacher also confirmed that the charge distribution will not be uniform like you said in that case. And the question I was talking about considered infinitely long conductor so charge density will be uniform in there. So that solves my problem :-)

John Rennie

10:08 AM

@Satwik cool :-)

Srijan M.T

Interesting. Thank you very much for today sir. I have solved a major chunk of my part today

John Rennie

@SrijanM.T :-)

Wolgwang

1:17 PM

> Two trains A and B of length 400 m each are moving on two parallel tracks with a
uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of
B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just
brushes past the driver of A, what was the original distance between them ?

Motion relative to train A

Intial velocity of train B=$0$

Initial distance between the train=$x$
$$800+x=\frac12(1)(50)^2\\x=450m$$

What did I do wrong?

Wolgwang

Displacement in nth second
$$S_n=un+\dfrac12an^2\\S_{n-1}=u(n-1)+\dfrac12a(n-1)^2$$
$$
\bbox[5px,border:2px solid red]
{S_n-S_{n-1}=u+\frac12a(2n-1)}$$
Isn't this dimensionally incorrect?

Satwik

1:56 PM

https://physics.stackexchange.com/questions/128172/the-dimensional-formula-of-distance-travelled-in-nth-second
Take a look at John Rennie's answer.

Wolgwang

Yes I observed it a few minutes ago :-D

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